Controlling A+B/B 0

Some numerical data on [math]\displaystyle{ |A+B/B_0| }[/math] source and also [math]\displaystyle{ \mathrm{Re} \frac{A+B}{B_0} }[/math] source, using a step size of 1 for [math]\displaystyle{ x }[/math], suggesting that this ratio tends to oscillate roughly between 0.5 and 3 for medium values of [math]\displaystyle{ x }[/math]:

range of [math]\displaystyle{ x }[/math]	minimum value	max value	average value	standard deviation	min real part	max real part
0-1000	0.179	4.074	1.219	0.782	-0.09	4.06
1000-2000	0.352	4.403	1.164	0.712	0.02	4.43
2000-3000	0.352	4.050	1.145	0.671	0.15	3.99
3000-4000	0.338	4.174	1.134	0.640	0.34	4.48
4000-5000	0.386	4.491	1.128	0.615	0.33	4.33
5000-6000	0.377	4.327	1.120	0.599	0.377	4.327
[math]\displaystyle{ 1-10^5 }[/math]	0.179	4.491	1.077	0.455	-0.09	4.48
[math]\displaystyle{ 10^5-2 \times 10^5 }[/math]	0.488	3.339	1.053	0.361	0.48	3.32
[math]\displaystyle{ 2 \times 10^5-3 \times 10^5 }[/math]	0.508	3.049	1.047	0.335	0.50	3.00
[math]\displaystyle{ 3 \times 10^5-4 \times 10^5 }[/math]	0.517	2.989	1.043	0.321	0.52	2.97
[math]\displaystyle{ 4 \times 10^5-5 \times 10^5 }[/math]	0.535	2.826	1.041	0.310	0.53	2.82
[math]\displaystyle{ 5 \times 10^5-6 \times 10^5 }[/math]	0.529	2.757	1.039	0.303	0.53	2.75
[math]\displaystyle{ 6 \times 10^5-7 \times 10^5 }[/math]	0.548	2.728	1.038	0.296	0.55	2.72

Here is a computation on the magnitude [math]\displaystyle{ |\frac{d}{dx}(B'/B'_0)| }[/math] of the derivative of [math]\displaystyle{ B'/B'_0 }[/math], sampled at steps of 1 in [math]\displaystyle{ x }[/math] source, together with a crude upper bound coming from the triangle inequality source, to give some indication of the oscillation:

range of [math]\displaystyle{ T=x/2 }[/math]	max value	average value	standard deviation	triangle inequality bound
0-1000	1.04	0.33	0.19
1000-2000	1.25	0.39	0.24
2000-3000	1.31	0.39	0.25
3000-4000	1.39	0.38	0.27
4000-5000	1.64	0.37	0.26
5000-6000	1.60	0.36	0.27
6000-7000	1.61	0.36	0.26
7000-8000	1.55	0.36	0.27
8000-9000	1.65	0.34	0.26
9000-10000	1.47	0.34	0.26
[math]\displaystyle{ 1-10^5 }[/math]	1.78	0.28	0.23	2.341
[math]\displaystyle{ 10^5-2 \times 10^5 }[/math]	1.66	0.22	0.18	2.299
[math]\displaystyle{ 2 \times 10^5-3 \times 10^5 }[/math]	1.55	0.20	0.17	2.195
[math]\displaystyle{ 3 \times 10^5-4 \times 10^5 }[/math]	1.53	0.19	0.16	2.109
[math]\displaystyle{ 4 \times 10^5-5 \times 10^5 }[/math]	1.31	0.18	0.15	2.039
[math]\displaystyle{ 5 \times 10^5-6 \times 10^5 }[/math]	1.34	0.18	0.14
[math]\displaystyle{ 6 \times 10^5-7 \times 10^5 }[/math]	1.33	0.17	0.14

In the toy case, we have

[math]\displaystyle{ \frac{|A^{toy}+B^{toy}|}{|B^{toy}_0|} \geq |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| }[/math]

where [math]\displaystyle{ b_n := \exp( \frac{t}{4} \log^2 n) }[/math], [math]\displaystyle{ a_n := (n/N)^{y} b_n }[/math], and [math]\displaystyle{ s := \frac{1+y+ix}{2} + \frac{t}{2} \log N + \frac{\pi i t}{8} }[/math]. For the effective approximation one has

[math]\displaystyle{ \frac{|A^{eff}+B^{eff}|}{|B^{eff}_0|} \geq |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| \quad (2.1) }[/math]

where now [math]\displaystyle{ b_n := \exp( \frac{t}{4} \log^2 n) }[/math], [math]\displaystyle{ s := \frac{1+y+ix}{2} + \frac{t}{2} \alpha_1(\frac{1+y+ix}{2}) }[/math], and

[math]\displaystyle{ a_n := |\frac{\exp( \frac{t}{4} \alpha_1(\frac{1-y+ix}{2})^2 ) H_{0,1}( \frac{1-y+ix}{2} )}{ \exp( \frac{t}{4} \alpha_1(\frac{1+y+ix}{2})^2 ) H_{0,1}( \frac{1+y+ix}{2} ) }| n^{y - \frac{t}{2} \alpha_1(\frac{1-y+ix}{2}) + \frac{t}{2} \alpha_1(\frac{1+y+ix}{2}) )} b_n. }[/math]

It is thus of interest to obtain lower bounds for expressions of the form

[math]\displaystyle{ |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| }[/math]

in situations where [math]\displaystyle{ b_1=1 }[/math] is expected to be a dominant term.

From the triangle inequality one obtains the lower bound

[math]\displaystyle{ |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| \geq 1 - |a_1| - \sum_{n=2}^N \frac{|a_n|+|b_n|}{n^\sigma} }[/math]

where [math]\displaystyle{ \sigma := \frac{1+y}{2} + \frac{t}{2} \log N }[/math] is the real part of [math]\displaystyle{ s }[/math]. There is a refinement:

Lemma 1 If [math]\displaystyle{ a_n,b_n }[/math] are real coefficients with [math]\displaystyle{ b_1 = 1 }[/math] and [math]\displaystyle{ 0 \leq a_1 \lt 1 }[/math] we have

[math]\displaystyle{ |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| \geq 1 - a_1 - \sum_{n=2}^N \frac{\max( |b_n-a_n|, \frac{1-a_1}{1+a_1} |b_n+a_n|)}{n^\sigma}. }[/math]

Proof By a continuity argument we may assume without loss of generality that the left-hand side is positive, then we may write it as

[math]\displaystyle{ |\sum_{n=1}^N \frac{b_n - e^{i\theta} a_n}{n^s}| }[/math]

for some phase [math]\displaystyle{ \theta }[/math]. By the triangle inequality, this is at least

[math]\displaystyle{ |1 - e^{i\theta} a_1| - \sum_{n=2}^N \frac{|b_n - e^{i\theta} a_n|}{n^\sigma}. }[/math]

We factor out [math]\displaystyle{ |1 - e^{i\theta} a_1| }[/math], which is at least [math]\displaystyle{ 1-a_1 }[/math], to obtain the lower bound

[math]\displaystyle{ (1-a_1) (1 - \sum_{n=2}^N \frac{|b_n - e^{i\theta} a_n| / |1 - e^{i\theta} a_1|}{n^\sigma}). }[/math]

By the cosine rule, we have

[math]\displaystyle{ (|b_n - e^{i\theta} a_n| / |1 - e^{i\theta} a_1|)^2 = \frac{b_n^2 + a_n^2 - 2 a_n b_n \cos \theta}{1 + a_1^2 -2 a_1 \cos \theta}. }[/math]

This is a fractional linear function of [math]\displaystyle{ \cos \theta }[/math] with no poles in the range [math]\displaystyle{ [-1,1] }[/math] of [math]\displaystyle{ \cos \theta }[/math]. Thus this function is monotone on this range and attains its maximum at either [math]\displaystyle{ \cos \theta=+1 }[/math] or [math]\displaystyle{ \cos \theta = -1 }[/math]. We conclude that

[math]\displaystyle{ \frac{|b_n - e^{i\theta} a_n|}{|1 - e^{i\theta} a_1|} \leq \max( \frac{|b_n-a_n|}{1-a_1}, \frac{|b_n+a_n|}{1+a_1} ) }[/math]

and the claim follows.

We can also mollify the [math]\displaystyle{ a_n,b_n }[/math]:

Lemma 2 If [math]\displaystyle{ \lambda_1,\dots,\lambda_D }[/math] are complex numbers, then

[math]\displaystyle{ |\sum_{d=1}^D \frac{\lambda_d}{d^s}| (|\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}|) = ( |\sum_{n=1}^{DN} \frac{\tilde b_n}{n^s}| - |\sum_{n=1}^{DN} \frac{\tilde a_n}{n^s}| ) }[/math]

where

[math]\displaystyle{ \tilde a_n := \sum_{d=1}^D 1_{n \leq dN} 1_{d|n} \lambda_d a_{n/d} }[/math]

[math]\displaystyle{ \tilde b_n := \sum_{d=1}^D 1_{n \leq dN} 1_{d|n} \lambda_d b_{n/d} }[/math]

Proof This is immediate from the Dirichlet convolution identities

[math]\displaystyle{ (\sum_{d=1}^D \frac{\lambda_d}{d^s}) \sum_{n=1}^N \frac{a_n}{n^s} = \sum_{n=1}^N \frac{\tilde a_n}{n^s} }[/math]

and

[math]\displaystyle{ (\sum_{d=1}^D \frac{\lambda_d}{d^s}) \sum_{n=1}^N \frac{b_n}{n^s} = \sum_{n=1}^N \frac{\tilde b_n}{n^s}. }[/math]

[math]\displaystyle{ \Box }[/math]

Combining the two lemmas, we see for instance that we can show [math]\displaystyle{ |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| \gt 0 }[/math] whenever can find [math]\displaystyle{ \lambda_1,\dots,\lambda_D }[/math] with [math]\displaystyle{ \lambda_1=1 }[/math] and

[math]\displaystyle{ \sum_{n=2}^N \frac{\max( \frac{|\tilde b_n-\tilde a_n|}{1-a_1}, \frac{|\tilde b_n+ \tilde a_n|}{1+a_1})}{n^\sigma} \lt 1. }[/math]

A usable choice of mollifier seems to be the Euler products

[math]\displaystyle{ \sum_{d=1}^D \frac{\lambda_d}{d^s} := \prod_{p \leq P} (1 - \frac{b_p}{p^s}) }[/math]

which are designed to kill off the first few [math]\displaystyle{ \tilde b_n }[/math] coefficients.

Analysing the toy model

With regards to the toy problem of showing [math]\displaystyle{ A^{toy}+B^{toy} }[/math] does not vanish, here are the least values of [math]\displaystyle{ N }[/math] for which this method works source source source source:

[math]\displaystyle{ P }[/math] in Euler product	[math]\displaystyle{ N }[/math] using triangle inequality	[math]\displaystyle{ N }[/math] using Lemma 1
1	1391	1080
2	478	341
3	322	220
5	282	192
7		180
11		176

Dropping the [math]\displaystyle{ \lambda_6 }[/math] term from the [math]\displaystyle{ P=3 }[/math] Euler factor worsens the 220 threshold slightly to 235 source.

Analysing the effective model

The differences between the toy model and the effective model are:

• The real part [math]\displaystyle{ \sigma }[/math] of [math]\displaystyle{ s }[/math] is now [math]\displaystyle{ \frac{1+y}{2} + \frac{t}{2} \mathrm{Re} \alpha_1(\frac{1+y+ix}{2}) }[/math] rather than [math]\displaystyle{ \frac{1+y}{2} + \frac{t}{2} \log N }[/math]. (The imaginary part of [math]\displaystyle{ s }[/math] also changes somewhat.)
• The coefficient [math]\displaystyle{ a_n }[/math] is now given by

[math]\displaystyle{ a_n = \lambda n^{y + \frac{t}{2} (\alpha_1(\frac{1+y+ix}{2}) - \alpha_1(\frac{1-y+ix}{2}))} b_n }[/math]

rather than [math]\displaystyle{ a_n = N^{-y} n^y b_n }[/math], where

[math]\displaystyle{ \lambda := |\frac{\exp( \frac{t}{4} \alpha_1(\frac{1-y+ix}{2})^2 H_{0,1}( \frac{1-y+ix}{2})}{\exp( \frac{t}{4} \alpha_1(\frac{1-y+ix}{2})^2 H_{0,1}( \frac{1-y+ix}{2})}|. }[/math]

Two complications arise here compared with the toy model: firstly, [math]\displaystyle{ \sigma,a_n }[/math] now depend on [math]\displaystyle{ x }[/math] and not just on [math]\displaystyle{ N }[/math], and secondly the [math]\displaystyle{ a_n }[/math] are not quite real-valued making it more difficult to apply Lemma 1.

However we have good estimates for [math]\displaystyle{ \sigma,a_n }[/math] that depend only on [math]\displaystyle{ N }[/math]. Note that

[math]\displaystyle{ 2\pi N^2 \leq T' \lt 2\pi (N+1)^2 }[/math]

and hence

[math]\displaystyle{ x_N \leq x \lt x_{N+1} }[/math]

where

[math]\displaystyle{ x_N := 4\pi N^2 - \frac{\pi t}{4}. }[/math]

To control [math]\displaystyle{ \sigma }[/math], it suffices to obtain lower bounds because our criteria (both the triangle inequality and Lemma 1) become harder to satisfy when [math]\displaystyle{ \sigma }[/math] decreases. We compute

[math]\displaystyle{ \sigma = \frac{1+y}{2} + \frac{t}{2} \mathrm{Re}(\frac{1}{1+y+ix} + \frac{2}{-1+y+ix} + \frac{1}{2} \log \frac{1+y+ix}{4\pi}) }[/math]

[math]\displaystyle{ = \frac{1+y}{2} + \frac{t}{2} (\frac{1+y}{(1+y)^2+x^2} + \frac{-2+2y}{(-1+y)^2+x^2} + \frac{1}{2} \log \frac{|1+y+ix|}{4\pi}) }[/math]

[math]\displaystyle{ \geq \frac{1+y}{2} + \frac{t}{2} (\frac{1+y}{(-1+y)^2+x^2} + \frac{-2+2y}{(-1+y)^2+x^2} + \frac{1}{2} \log \frac{x}{4\pi}) }[/math]

[math]\displaystyle{ \geq \frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{(-1+y)^2+x^2} + \log N) }[/math]

[math]\displaystyle{ \geq \frac{1+y}{2} + \frac{t}{2} \log N }[/math]

assuming that [math]\displaystyle{ y \geq 1/3 }[/math]. Hence we can actually just use the same value of [math]\displaystyle{ \sigma }[/math] as in the toy case.

Next we control [math]\displaystyle{ \lambda }[/math]. Note that we can increase [math]\displaystyle{ \lambda }[/math] (thus multiplying [math]\displaystyle{ \sum_{n=1}^N \frac{a_n}{n^s} }[/math] by a quantity greater than 1) without affecting (2.1), so we just need upper bounds on [math]\displaystyle{ \lambda }[/math]. We may factor

[math]\displaystyle{ \lambda = \exp( \frac{t}{4} \mathrm{Re} (\alpha_1(\frac{1-y+ix}{2})^2 - \alpha_1(\frac{1+y+ix}{2})^2) + \mathrm{Re}( f(\frac{1-y+ix}{2}) - f(\frac{1+y+ix}{2} ) ) }[/math]

where

[math]\displaystyle{ f(s) := -\frac{s}{2} \log \pi + (\frac{s}{2} - \frac{1}{2}) \log \frac{s}{2} - \frac{s}{2}. }[/math]

By the mean value theorem, we have

[math]\displaystyle{ \mathrm{Re} (\alpha_1(\frac{1-y+ix}{2})^2 - \alpha_1(\frac{1+y+ix}{2})^2) = -2 y \alpha_1(s') \alpha'_1(s') }[/math]

for some [math]\displaystyle{ s_1 }[/math] between [math]\displaystyle{ \frac{1-y+ix}{2} }[/math] and [math]\displaystyle{ \frac{1+iy}{2} }[/math]. We have

[math]\displaystyle{ \alpha_1(s_1) = \frac{1}{2s_1} + \frac{1}{s_1-1} + \frac{1}{2} \log \frac{s_1}{2\pi} }[/math]

[math]\displaystyle{ = O_{\leq}(\frac{1}{x}) + O_{\leq}(\frac{1}{x/2}) + \frac{1}{2} \log \frac{|s_1|}{2\pi} + O_{\leq}(\frac{\pi}{4}) }[/math]

[math]\displaystyle{ = O_{\leq}( \frac{\pi}{4} + \frac{3}{x_N}) + \frac{1}{2} O_{\leq}^{\mathbf{R}}( \log \frac{|1+y+ix_{N+1}|}{4\pi} ) }[/math]

and

[math]\displaystyle{ \alpha'_1(s_1) = -\frac{1}{2s_1^2} + \frac{1}{(s_1-1)^2} + \frac{1}{2s_1} }[/math]

[math]\displaystyle{ = O_{\leq}(\frac{1}{x^2/2}) + O_{\leq}(\frac{1}{x^2/4}) + \frac{1}{2s_1} }[/math]

[math]\displaystyle{ = O_{\leq}(\frac{6}{x_N^2}) + \frac{1}{2s_1} }[/math]

[math]\displaystyle{ = O_{\leq}(\frac{6}{x_N^2}) + O_{\leq}( \frac{1}{x_N} ). }[/math]

Thus one has

[math]\displaystyle{ \mathrm{Re} (\alpha_1(\frac{1-y+ix}{2})^2 - \alpha_1(\frac{1+y+ix}{2})^2) = 2y O_{\leq}( (\frac{\pi}{4} + \frac{3}{x_N}) (\frac{1}{x_N} + \frac{6}{x_N^2}) ) }[/math]

[math]\displaystyle{ + 2y O_{\leq}( \log \frac{|1+y+ix_{N+1}|}{4\pi} (\frac{6}{x_N^2} + |\mathrm{Re} \frac{1}{2s'}|) ) }[/math]

Now we have

[math]\displaystyle{ \mathrm{Re} \frac{1}{2s'} = \frac{\mathrm{Re}(s')}{2|s'|^2} }[/math]

[math]\displaystyle{ \leq \frac{1+y}{x^2} }[/math]

[math]\displaystyle{ \leq \frac{1+y}{x_N^2}; }[/math]

also

[math]\displaystyle{ (\frac{\pi}{4} + \frac{3}{x_N}) (\frac{1}{x_N} + \frac{6}{x_N^2}) \leq \frac{\pi}{4} (1 + \frac{12/\pi}{x_N}) \frac{1}{x_N-6} }[/math]

[math]\displaystyle{ \leq \frac{\pi}{4} ( \frac{1}{x_N-6} + \frac{12/\pi}{(x_N-6)^2} ) }[/math]

[math]\displaystyle{ \leq \frac{\pi}{4} \frac{1}{x_N - 6 - 12/\pi}. }[/math]

We conclude that

[math]\displaystyle{ \mathrm{Re} (\alpha_1(\frac{1-y+ix}{2})^2 - \alpha_1(\frac{1+y+ix}{2})^2) = O_{\leq}(\frac{\pi y}{2 (x_N - 6 - 12/\pi)} + \frac{2y(7+y)}{x_N^2} \log \frac{|1+y+ix_{N+1}|}{4\pi}). }[/math]

In a similar vein, from the mean value theorem we have

[math]\displaystyle{ \mathrm{Re}( f(\frac{1-y+ix}{2}) - f(\frac{1+y+ix}{2} ) = -y \mathrm{Re} f'(s_2) }[/math]

for some [math]\displaystyle{ s_2 }[/math] between [math]\displaystyle{ \frac{1-y+ix}{2} }[/math] and [math]\displaystyle{ \frac{1+y+ix}{2} }[/math]. We have

[math]\displaystyle{ \mathrm{Re} f'(s_2) = -\frac{1}{2} \log \pi + \frac{1}{2} \log \frac{|s_2|}{2} - \mathrm{Re} \frac{1}{2s_2} }[/math]

[math]\displaystyle{ = \frac{1}{2} \log \frac{|s_2|}{2\pi} + O_{\leq}(\frac{\mathrm{Re}(s_2)}{2|s_2|^2}) }[/math]

[math]\displaystyle{ \geq \log N + O_{\leq}(\frac{1+y}{x^2}) }[/math]

[math]\displaystyle{ \geq \log N + O_{\leq}(\frac{1+y}{x_N^2}) }[/math]

and thus

[math]\displaystyle{ \lambda \leq N^{-y} \exp( \frac{\pi y}{2 (x_N - 6 - 12/\pi)} + \frac{2y(7+y)}{x_N^2} \log \frac{|1+y+ix_{N+1}|}{4\pi} + \frac{y(1+y)}{x_N^2} ) }[/math]

[math]\displaystyle{ \leq e^\delta N^{-y} }[/math]

where

[math]\displaystyle{ \delta := \frac{\pi y}{2 (x_N - 6 - \frac{14+2y}{\pi})} + \frac{2y(7+y)}{x_N^2} \log \frac{|1+y+ix_{N+1}|}{4\pi} }[/math]

Asymptotically we have

[math]\displaystyle{ \delta = \frac{\pi y}{2 x_N} + O( \frac{\log x_N}{x_N^2} ) = O( \frac{1}{x_N} ). }[/math]

Now we control [math]\displaystyle{ \alpha_1(\frac{1+y+ix}{2}) - \alpha_1(\frac{1-y+ix}{2}) }[/math]. By the mean-value theorem we have

[math]\displaystyle{ \alpha_1(\frac{1+y+ix}{2}) - \alpha_1(\frac{1-y+ix}{2}) = O_{\leq}( y |\alpha'_1(s_3)|) }[/math]

for some [math]\displaystyle{ s_3 }[/math] between [math]\displaystyle{ \frac{1+y+ix}{2} }[/math] and [math]\displaystyle{ \frac{1-y+ix}{2} }[/math]. As before we have

[math]\displaystyle{ \alpha'_1(s_3) = -\frac{1}{2s_3^2} - \frac{1}{(s_3-1)^2} + \frac{1}{2s_3} }[/math]

[math]\displaystyle{ = O_{\leq}( \frac{1}{x^2/2} + \frac{1}{x^2/4} + \frac{1}{x} ) }[/math]

[math]\displaystyle{ = O_{\leq}( \frac{1}{x_N} + \frac{6}{x_N^2} ) }[/math]

[math]\displaystyle{ = O_{\leq}( \frac{1}{x_N-6} ). }[/math]

We conclude that (after replacing [math]\displaystyle{ \lambda }[/math] with [math]\displaystyle{ e^\delta N^{-y} }[/math])

[math]\displaystyle{ a_n = (n/N)^y \exp( \delta + O_{\leq}( \frac{t y \log n}{2(x_N-6)} ) ) b_n. }[/math]

The triangle inequality argument will thus give [math]\displaystyle{ A^{eff}+B^{eff} }[/math] non-zero as long as

[math]\displaystyle{ \sum_{n=1}^N (1 + (n/N)^y \exp( \delta + \frac{t y \log n}{2(x_N-6)} ) ) \frac{b_n}{n^\sigma} \lt 2. }[/math]

The situation with using Lemma 1 is a bit more complicated because [math]\displaystyle{ a_n }[/math] is not quite real. We can write [math]\displaystyle{ a_n = e^\delta a_n^{toy} + O_{\leq}( e_n ) }[/math] where

[math]\displaystyle{ a_n^{toy} := (n/N)^y b_n }[/math]

and

[math]\displaystyle{ e_n := e^\delta (n/N)^y (\exp( \frac{t y \log n}{2(x_N-6)} ) - 1) b_n }[/math]

and then by Lemma 1 and the triangle inequality we can make [math]\displaystyle{ A^{eff}+B^{eff} }[/math] non-zero as long as

[math]\displaystyle{ a_1^{toy} + \sum_{n=2}^N \frac{\max( |b_n-a_n^{toy}|, \frac{1-a_1^{toy}}{1+a_1^{toy}} |b_n + a_n^{toy}|}{n^\sigma} + \sum_{n=1}^N \frac{e_n}{n^\sigma} \lt 1. }[/math]